## 每週問題 November 19, 2012

If $\mathbf{b}=(1,3,0,2)^T$ and the general solution to $A\mathbf{x}=\mathbf{b}$ is

$\mathbf{x}=\left[\!\!\begin{array}{r} 1\\ 1\\ -1 \end{array}\!\!\right]+\alpha\begin{bmatrix} 1\\ 2\\ 0 \end{bmatrix}+\beta\left[\!\!\begin{array}{r} 2\\ 2\\ -1 \end{array}\!\!\right],$

where $\alpha$ and $\beta$ are free parameters, determine the matrix $A$.

$A\left[\!\!\begin{array}{r} 1\\ 1\\ -1 \end{array}\!\!\right]=\begin{bmatrix} 1\\ 3\\ 0\\ 2 \end{bmatrix},~~ A\begin{bmatrix} 1\\ 2\\ 0 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix},~~ A\left[\!\!\begin{array}{r} 2\\ 2\\ -1 \end{array}\!\!\right]=\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix}$

$A\left[\!\!\begin{array}{rrr} 1&1&2\\ 1&2&2\\ -1&0&-1 \end{array}\!\!\right] =\begin{bmatrix} 1&0&0\\ 3&0&0\\ 0&0&0\\ 2&0&0 \end{bmatrix}$

\begin{aligned} A&=\begin{bmatrix} 1&0&0\\ 3&0&0\\ 0&0&0\\ 2&0&0 \end{bmatrix}\left[\!\!\begin{array}{rrr} 1&1&2\\ 1&2&2\\ -1&0&-1 \end{array}\!\!\right]^{-1}=\begin{bmatrix} 1&0&0\\ 3&0&0\\ 0&0&0\\ 2&0&0 \end{bmatrix}\left[\!\!\begin{array}{rrr} -2&1&-2\\ -1&1&0\\ 2&-1&1 \end{array}\!\!\right]\\ &=\left[\!\!\begin{array}{rrr} -2&1&-2\\ -6&3&-6\\ 0&0&0\\ -4&2&-4 \end{array}\!\!\right].\end{aligned}

PowSol-Nov-19-12

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